Download Assignment 1 with Solution | Applied Thermodynamics | M E 320 and more Assignments Thermodynamics in PDF only on Docsity!
1.4) An object has a mass of 10 lb. Determine its weight, in lbf, at a location where the acceleration of gravity is 31.0 ft/s^2.
1 ๐๐ = 4.536^ ๐๐
๐ ^2 ๐ฅ^
3.2808 ๐๐ก = 9.5^ ๐/๐
2
๐ ^2 )
weight = 43.092 N โ43.1 N
๐ค๐๐๐๐๐ก = 43. 1 ๐ ๐ฅ 0.^224811 ๐ ๐๐๐= 9. 67 ๐๐๐ ANS
1.31) A gas contained within a piston-cylinder assembly undergoes three processes in series:
Process 1-2 : Compression with pV =constant from p 1 = 1 bar, V 1 = 1.0 m^3 to V 2 = 0.2 m^3
Process 2-3 : Constant-=pressure expansion to V 3 = 1.0 m^3
Process 3-1 : Constant volume
Sketch the processes in series on a p-V diagram labeled with pressure and volume values at each numbered state.
First determine the pressures and volumes at each of the states (only P 2 unknown):
P 2 =?
P 1 V 1 = P 2 V 2 (Process 1-2 has pV = constant)
๐ 2 =^
1 ๐๐๐ (1.0 ๐^3 )
(0.2 ๐^3 ) = 5^ ๐๐๐๐
State Pressure (bars) Volume (m^3 ) 1 1 1 2 5 0. 3 5 1
Sketch the process and label the states and their pV values (labeled as (p,V)):
1.34) The absolute pressure inside a tank is 0.4 bars, and the surroundings atmospheric pressure is 98 kPa. What reading would a Bourdon gage mounted in the tank wall give, in kPa? Is this a gage or vacuum reading?
First convert bars to kPa:
1 ๐ฅ 105 ๐/๐^2
1 ๐/๐^2
A Bourdon gage measures the difference between the absolute internal pressure and the absolute outside pressure. In this case the tank pressure and the atmospheric pressure. Since the atmospheric pressure is greater than the tank pressure, the measured pressure will be a vacuum pressure:
2.1) A baseball has a mass of 0.3 lb. What is the kinetic energy relative to home plate of a 94 mile per hour fastball, in Btu?
2
Converting from English to SI units:
2.2046 ๐๐ = 0.136^ ๐๐
1 ๐๐๐ = 151.27^ ๐๐/๐
Solving for the kinetic energy:
๐๐^2
๐^2
(1๐ฅ 103 ๐)^2
(1 ๐๐)^2
(1 ๐)^2
(60 ๐๐๐)^2
(1 ๐๐๐)^2
(60 ๐ )^2
๐^2
๐ ^2
๐^2
๐ ^2 ๐ฅ^
๐๐ ๐/๐ ^2 = 120.83^ ๐^ ๐^ ๐ฅ^
1 ๐ ๐ = 120.83^ ๐ฝ
Converting final answer into Btu yields:
1000 ๐ฝ ๐ฅ^
1 ๐๐ฝ =^0.^1145 ๐ต๐ก๐ข
๐พ๐ธ = 0.1145 ๐ต๐ก๐ข ANS
2.20) The drag force, Fd, imposed by the surrounding air on a vehicle moving with velocity V is given by:
F (^) d = Cd A^12 ฯV^2
where Cd is a constant called the drag coefficient, A is the projected frontal area of the vehicle, and ฯ is the air density. Determine the power, in kW, required to overcome aerodynamic drag for a truck moving at 110 km/h, if Cd = 0.65, A = 10m^2 , and ฯ = 1.1 kg/m^3.
Solving for the draq force Fd yields:
F (^) d = Cd A^12 ฯV^2
๐น๐ = 0.65 10 ๐^2 0.5 1.
๐^3 (^
2
๐^2 ๐๐^2
๐^3 ๐^2 = 4.326๐ฅ^10
2 ๐ ๐^2
๐น๐ = 4.326๐ฅ 104 ๐๐
๐๐^2
๐ ๐^2
(1000 ๐)^2
(1๐๐)^2
(1 ๐)^2
(3600๐ )^2
๐๐ ๐/๐ ^2
Solving for the power:
๐ = ๐น โ ๐ = ๐น ๐ cos ๐
Since the drag force acts directly opposite the velocity, ฮธ = 180
๐๐ = 3.338๐ฅ 103 ๐
cos 180 = 3.338๐ฅ 103 110
๐ ๐ฅ^
1 ๐๐ ๐ฅ^
3600 ๐ ๐ฅ^
1 ๐๐ ๐ฅ^
1 ๐ฝ/๐ =^ โ1.02๐ฅ^10
The power required to overcome this aerodynamic drag force is the opposite of the value calculated above, therefore:
๐ = โ๐๐ = +1.02๐ฅ 105 ๐ ๐ฅ
๐ = 102 ๐๐ ANS
